%I #4 Sep 27 2014 19:03:02
%S 1,3,5,6,8,10,11,13,14,16,18,19,21,23,24,26,28,29,31,33,34,36,38,39,
%T 41,43,44,46,48,49,51,53,54,56,58,59,61,63,65,66,68,70,71,73,75,76,78,
%U 80,82,83,85,87,88,90,92,94,95,97,99,100,102,104,106,107
%N Numbers k such that A247911(k+1) = A247911(k) + 1.
%C Complement of A247912.
%H Clark Kimberling, <a href="/A247913/b247913.txt">Table of n, a(n) for n = 1..500</a>
%e A247911(n+1)  A247911(n) = (1,0,1,0,1,1,0,1,0,1,1,0,1,1,0,1,0,1,1,0,...), and a(n) is the position of the nth 1.
%t $RecursionLimit = 1000; $MaxExtraPrecision = 1000;
%t z = 300; u[1] = 0; u[2] = 1; u[n_] := u[n] = u[n  1] + u[n  2]/(n  2);
%t f[n_] := f[n] = Select[Range[z], (2 # + 1)/u[2 # + 1]  E < n^n &, 1];
%t u = Flatten[Table[f[n], {n, 1, z}]] (* A247911 *)
%t w = Differences[u]
%t Flatten[Position[w, 0]] (* A247912 *)
%t Flatten[Position[w, 1]] (* A247913 *)
%Y Cf. A247911, A247912, A247908, A247914.
%K nonn,easy
%O 1,2
%A _Clark Kimberling_, Sep 27 2014
